package leetcode.array;

/**
 * 给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。
 * <p>
 * 请必须使用时间复杂度为 O(log n) 的算法。
 * <p>
 * 示例 1:
 * <p>
 * 输入: nums = [1,3,5,6], target = 5
 * 输出: 2
 * 示例 2:
 * <p>
 * 输入: nums = [1,3,5,6], target = 2
 * 输出: 1
 * 示例 3:
 * <p>
 * 输入: nums = [1,3,5,6], target = 7
 * 输出: 4
 */
// 注意 时间复杂度为 log2N的算法就是二分查找
public class _35 {

    public static void main(String[] args) {
        int[] arr = new int[]{1, 3, 5, 6};
        int i = searchInsertStd(arr, 7);
        System.out.println(i);
    }

    public static int searchInsert(int[] nums, int target) {
        if (target <= nums[0]) return 0;
        if (target >= nums[nums.length - 1]) {
            return nums.length;
        }
        int left = 0;
        int right = nums.length - 1;
        while (right - left != 1 && left <= right) {
            int middle = (right - left) / 2 + left;

            if (target > nums[middle]) {
                left = middle;
            } else if (target < nums[middle]) {
                right = middle;
            } else {
                return middle;
            }
        }
        return left + 1;

    }



    public static int searchInsertStd(int[] nums, int target) {
        int n = nums.length;
        int left = 0, right = n - 1, ans = n;
        while (left <= right) {
            int mid = ((right - left) >> 1) + left;
            if (target <= nums[mid]) {
                ans = mid;
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return ans;
    }

}
